Find $\lim_{\theta\to\scriptsize\dfrac{\pi}{2}}\tan^2(\theta)[1-\sin(\theta)]$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{2}$ (Choice C) C $-2$ (Choice D) D The limit doesn't exist
Explanation: Substituting $\theta=\dfrac{\pi}{2}$ into $\tan^2(\theta)[1-\sin(\theta)]$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\cos^2(\theta)$ in our expression, because $\tan^2(\theta)=\dfrac{\sin^2(\theta)}{\cos^2({\theta})}$, let's rewrite it using the Pythagorean identity, $\sin^2(\theta)+\cos^2(\theta)=1$ : $\begin{aligned} &\phantom{=}\tan^2(\theta)[1-\sin(\theta)] \\\\ &=\dfrac{\sin^2(\theta)[1-\sin(\theta)]}{\cos^2({\theta})} \gray{\text{Definition of tangent}} \\\\ &=\dfrac{\sin^2(\theta)[1-\sin(\theta)]}{(1-\sin^2(\theta))} \gray{\text{The Pythagorean identity}} \\\\ &=\dfrac{\sin^2(\theta)[1-\sin(\theta)]}{(1-\sin(\theta))(1+\sin(\theta))} \gray{\text{Diff. of squares}} \\\\ &=\dfrac{\sin^2(\theta)[\cancel{1-\sin(\theta)}]}{\cancel{(1-\sin(\theta))}(1+\sin(\theta))} \gray{\text{Cancel common factors}} \\\\ &=\dfrac{\sin^2(\theta)}{1+\sin(\theta)}\text{, for }\theta\neq \{...,- \dfrac{7\pi}{2}, - \dfrac{3\pi}{2}, \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},...\} \end{aligned}$ This means that the two expressions have the same value for all $\theta $ -values (in their domains) except for $(2k+1)\dfrac{\pi}{2}$ for any integer $k$, and specifically $\dfrac{\pi}{2}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\tan^2(\theta)[1-\sin(\theta)]=\dfrac{\sin^2(\theta)}{1+\sin(\theta)}$ for all $\theta$ -values in the interval $(-\pi,\pi)$ except for $\theta=\dfrac{\pi}{2}$. Therefore, $\lim_{\theta\to \scriptsize\dfrac{\pi}{2}}\tan^2(\theta)[1-\sin(\theta)]=\lim_{\theta\to \scriptsize\dfrac{\pi}{2}}\dfrac{\sin^2(\theta)}{1+\sin(\theta)}=\dfrac12$ (The last limit was found using direct substitution.) In conclusion, $\lim_{\theta\to\scriptsize\dfrac{\pi}{2}}\tan^2(\theta)[1-\sin(\theta)]=\dfrac12$.